In the linked article, John D. Kubiatowicz, a professor of computer science at the University of California, Berkeley says that for a kindle with 4 Gigabyte memory, this would mean a rough increase in weight of 10–18 grams.

For your situation of a 60GB phone, this would mean (60Gb/4Gb) * (10-18 grams) = 1.5 × 10-17 grams. So certainly less than what you would be able to feel.

While I originally thought this would be because the storage would use extra electrons, the article/Prof. Kubiatowicz say otherwise:

"Although the total number of electrons in the memory does not change as the stored data changes,” Dr. Kubiatowicz said, the trapped ones have a higher energy than the untrapped ones. A conservative estimate of the difference would be 10–15 joules per bit.

"As the equation E=mc2 makes clear, this energy is equivalent to mass and will have weight."

It is my understanding that most/all mobile phones use flash memory for storage so the linked article about flash memory in Kindles would be applicable here. Please let me know (or just delete this outright) if I am mistaken.

As far as I know, it's applicable. But, as the article states, that would be magnitudes less than just fluctuations due to battery level ("one hundred-millionth"). And at that small scale, we wouldn't be able to measure it anyways.

It's not that you're measuring extra electrons, in fact, the electrons DON'T move from the memory to the battery or vice versa. The number of electrons in the memory stays the same, but the transistors holding a "high" value have trapped the electrons in a higher energy state and, therefore, provide more mass.

But where does the energy come from? Isnt it coming from some other electrons, ie the battery, and passing it along? So while the amount of electrons stays the same they still got something from the battery, or not?

The electrons just flow from one terminal of the battery to the other, and do work in the process. All electrons that flow out of the battery just flow right back in the other side in a lower energy state. The energy comes from the change in energy state after interacting with the circuit.

I think in a closed system you are right. There* could not be a change in mass/energy because of conservation laws. In a non-isolated system the process of writing the data would lose heat energy, more than the data storage would add but that energy could be supplied from outside.A proper comparison I think would be:

Two identical phones, A and B with 60GB free each.Neither has a battery and both are switched off.Both are simultaneously plugged in to an external power source and switched on.A writes data to its 60GB. B does not.Both are simultaneously switched off and disconnected from the power.Both are given time to equalise their temperatures.

I think then we would expect A to have a slightly greater mass as it has drawn and retained more power while it was processing and storing the data.

Sure, but the electrons transitioning to a higher energy state inside the transistors are using energy provided by the battery. The phone itself would not get heavier, because any increase in mass of the storage device would be offset by the decrease in mass of the battery.

To add to this most storage devices don't have their unused space kept at all 0's, the unused space is usually random 1's and 0's. It's entirely possible that by saving files to your phone you could end up with more 0's and less 1's than you started with and your phone could end up "lighter".

On flash devices empty blocks are usually erased in the background when possible (see TRIM), since erasure is an expensive operation and doing it ahead of time means you'll get less hiccups later on when you want to write a bunch of new data quickly.

No, since it would have to be erased eventually anyway. Unlike traditional magnetic media, you must erase before you can write on Flash. And Erasing is slow, so it makes sense to get it done ahead of time.

One limitation of flash memory is that, although it can be read or programmed a byte or a word at a time in a random access fashion, it can be erased only a block at a time. This generally sets all bits in the block to 1. Starting with a freshly erased block, any location within that block can be programmed. However, once a bit has been set to 0, only by erasing the entire block can it be changed back to 1.

A write (one or more bits to 0) and reset (whole block to 1) are completely different operations.

Yes--ish. The problem is that the hardware design doesn't allow for turning a single 0 into a 1. If you wanted to do that, you would need to fry the entire block containing that bit to be 1's, then put back the rest of the 0's.

Also, IIRC only the reset direction stresses and puts wear on the cells.

You can only flip single bits to 0. To get 1s, you have to flip the whole cell to 1s then flip whatever needed to be 0 back to 0. The erase (flip to 1s) takes forever (relatively speaking). You're much better off erasing when you have down time. Plus, write ware hasn't been an issue with flash store for a long time. Your drive will die of something else or you'll replace it because it's uselessly small long before you wear the drive from writes.

I don't believe so. Because the act of deleting is more related to the file system used than the type of storage media, it's pretty much the same, but ultimately depends on what you define as "delete". Typical file systems (NTFS, etc...) don't delete in the sense of flipping bits to 0.

* Heavier. Flash memory uses floating-gate transistors. When a charge is stored in the floating-gate, the transistor does not conduct when voltage is applied to it, and thus it reads as a logical 0. An uncharged FG make the transistor conductive, reading a logical 1. Therefore erased flash memory set to all 1s would consist of uncharged floating-gates, and would be the lightest state.

Well, it depends what the "favored", or ground state, position is. Basically what it would decay to given infinite amount of time.

So if the answer is in the middle, the natural state would be roughly 50% randomly distributed of both and then all 1's and all 0's would both weigh more than a randomly distributed state, and they would weigh the same.

However, if 0's are the ground state, and 1's are some excited state, then given infinite amount of time the system would tend towards mostly 0's, and being filled with all 1's would be the heaviest state.

With magnetic media, I believe both states are roughly equally probable, because you're flipping a magnetic charge which has two sides, like rotating a ball so the North Pole points up or down, thus it would be more like the former situation, randomly distributed.

Flash media uses an electric charge, electrons specifically, which is more of a 'charged or not charged' state, so in that case it would be more like the latter situation, where all 1's would weigh the most, and the closer you get to all 0's the less mass.

Assuming that all these bits in an empty four-gigabyte Kindle are in a lower energy state and that half have a higher energy in a full Kindle, this translates to an energy difference of 1.7 times 10–5 joules, Dr. Kubiatowicz calculated. Plugging this into Einstein’s equation yields his rough estimate of 10–18 grams.

The article does the calculation assuming that half the bits are set to 1.

Well if a bit 10-15 joules as quoted by /u/johnmon, then that means the amount of mass would be 1.11265 x 10-32 kg. (Alternatively, if a bit is supposed to be the mass of an electron, wolframalpha tells me that's 0.9109383 x 10-32 so probably well within in a rounding error as the joules were heavily rounded I assume.)

So for the hell of it, I'm using 1x10-32 kg = 1 bit (or electron).

Now the mass of a strawberry. Wolframalpha tells me 12g or 0.012 kg. So how many bits are in a strawberry?

Wolframalpha estimates the deep web for 3 years ago to be 91,000 Terrabytes. And the surface web to be 170 TB. 91,170 TB. But for the hell of it, let's say Moore's Law applies to data creation on the web and we'll raise it to 23/2 power for doubling every two years, but it's three years later. We could kick this to being nearly 260,000 TB.

No. Wolfram was dumb for a moment (two moments, as explained in comments below). Searching Internet Size, the 2014 estimate is 1x106 Exabytes which is 1 Yottabyte.

That's a few magnitudes short of a strawberry in terms of raw data. The 1.5 x 1029 bytes is 1.5 x 105 or 175,000 Yottabytes. A strawberry is still a lot more.

However! If we mean to look at just the DNA inside of a Strawberry, that's a different question to tackle. I imagine it brings it into the realm too low, given how massive DNA is, requiring many atoms to make one bit.

Oh, but that's also true. I treated my numbers to be fully electrons, but if we convert to each atom being a bit, that would help bring it into a similar magnitude.

Back in the old days of the web, everyone was supposed to run servers and we would both provide content and consume content. The media companies bought ISPs and blocked ports and most internet users are only consumers. But I member.

...but those electrons would still all be there if you erased the internet (by writing over it, on every hard drive etc., with all zeroes). The weight of the energy differential between "the internet" and "the internet overwritten with all zeroes" would be much smaller than that strawberry.

Then let's assume all of it is held on 6TB hard drives which weigh about 1.7lbs a piece.

The total weight would be 3,289,389,216lbs (1,492,041,850kg). Or 23 aircraft carriers, more than the entire world has in service.

And that's not even counting the servers that hold those hard drives, or the racks, or the datacenters built to house them, or the tons of equipment to keep that datacenter cool and powered. It really becomes impossible to calculate very quickly.

I'm really struggling to accept this answer. We're talking about non-volatile memory here, so it doesn't have a charge to retain state. So what are we talking about when we say e? What energy does that refer to?

The explanation in that link was completely inapplicable to the question. It basically claims that 0-bits are denser than 1-bits at the transistor level. But that doesn't have anything at all to do with data storage, where the bits are effectively equal in count.

This is talking about floating-gate flash storage. By overvolting the transistor electrons are tunneled through the substrate and get stuck on the floating gate. They don't have enough potential to tunnel back through the substrate on their own from there, but they do have some potential (against the transistor drain), so there is some potential energy that ever so slightly bends spacetime.

Yes and correct. Also, a closed system that isn't gaining energy isn't gaining mass. The whole phone can't magically gain mass from nothing. Now, you could flip the bits, and charge it and potentially have the phone gain weight (no longer a closed system).

And this effect will outweigh the effect of the data-mass equivalency. Pun intended. Solving for resting mass using a decent approximation of the energy of a large smartphone gives 3000 mAh * 3.7 V / c^2, a mass many orders of magnitude larger.

The only problem is that your scale would have to be so sensitive that electrons would cause it to fluctuate, so it would be really hard to isolate from the noise. You'd likely have to move massive amounts of data (yottabytes) to be able to detect it with any certainty

This is the wrong answer as the assumption of the article is that all bits are in a lower energy state to start with. That is not how data storage works. We don't know the initial state of the storage. There is always 100% "data" on the storage the only thing that changes is the readability and perceived usefulness of that data. A byte may be 10011011 in storage as unreadable gibberish, but when it is written to with a filesystem it may be 01001000 which would be higher or lower in energy depending on the file system design.

Can you elaborate on the relationship between file system design and energy level changes? To me and my unknowledgeable brain, file system design sounds like a software thing, whereas electron energy states seem like a hardware thing.

Just referring to what the software uses as a state of 1. The high energy state could be used as 1 or the low enee gy state could be used as 1. So depending on how the software uses the availalable storage a 1 could be either the high or low energy state. Think of a relay switch. When programming the "open" state of the switch you can choose either the high energy gate or low energy gate. So "on" could be 1 or 0 for the relay.

Interesting.. but it wouldn't be precise to call it an increase in "mass". Rather, Einstein's equations equate the geometry of space with energy, mass, & momentum. So presumably the additional energy would provide a very, very small contribution to the curvature of space-time that could be achieved by a mass of ~10-18 grams.

1.5x10-17 grams is as far from nano-level as grams themselves are. A nanogram is 1x10-9. So that's 100,000,000 times smaller than a nanogram. That’s millions of times lighter than bacteria. That’s closer to the magnitude of a proton’s mass than a paperclip’s mass.

You're not wrong - the idea behind this is pretty contrived. You have to take it as a hypothetical of the difference in weight between phones with and without memory being full, but all other conditions exactly the same. It doesn't really mean much at that point that they don't weigh the same.

Obviously in reality, no matter what is happening in memory the phone as a whole system must be consistently losing energy/mass as the battery discharges (unless something is heating it up). Each time I've read about this it seems a little silly, because it's similar to saying, "When I drip a couple drops out of this enormous bucket into these tiny cups, the cups have some water in them and the big bucket has slightly less. And I lose water entirely in the process because an overwhelming majority misses the cups." Like, yeah ?

Correct me if I'm wrong but wouldn't this only be for devices with solid state storage? Of course there's still the electrons in the RAM but a device with a HDD doesn't have "electronic" memory. Just magnetized sectors on a platter.

This is nonsense as it assumes some sort of bit filling as data is written to the phone. In reality, the actual bits are probably pretty high entropy at all times (about 50/50 zeros, ones). Just because your controller/firmware considers bytes unused doesn't mean they aren't something like 10010101 physically. Even in NAND flash, I don't think unallocated bytes are zeroed out as that seems like an unnecessary write operation that would decrease the life of the memory. I can't think of a single storage medium where this would not be the case.

It's true that the high energy cells (with higher voltage electrons) will have fractionally more mass, but something I haven't seen mentioned is that having more data on a modern phone won't actually cause more cells to be turned on (on average), even if that data is just a bunch of '1's (eg. very bright/white bitmap images).

The capacitors in flash storage are quite sensitive to performance degradation depending on how they're charged and discharged, so all modern SSDs encrypt the data before storing it to 'whiten' the data and make the bits uniformly random. This means that changing a few bits over and over again (eg. frequently saving a document you're working-on) won't always charge the same capacitors, so they won't wear out quickly. It also means that filling up your device with data will cause a random ~50% of cells to be charged, even if the bits in your data are almost all '1's

What if the phone was a quantum phone. Would it still gain weight based on energy or since all electrons are already dispositioned due to quantum mechanics would it be a finite weight regardless of actual storage taken?

I would have thought the weight stayed the same. That full space weighed the same as empty space. Say empty space is all 0's, and the data converts some of them to 1's. There's still an equivalent number of digits

I guess the act of changing some of those 0's to 1's has some energy transference which impacts weight?

It depends on how you think about mass. Trapped energy exhibits all the properties of mass: that is, it has inertia and an increased gravitational field.

As a thought experiment, imagine you could construct a box whose insides are lined with ideal perfect mirrors (completely lossless reflection). If you were able to inject visible light into such a box, say a few trillion photons, the photons would bounce back and forth inside the box. Moving the box, you'd find it would have more inertia than a box absent photons. This is because you have to actually impart momentum to the photons inside the box, as well as the box itself, to move it (yes, photons can have momentum).

Similarly, there is a theoretical object that physics says is definitely possible called a Kugelblitz -- if you shine ridiculously powerful lasers onto a single point, all the photons in that area are so dense that their gravitational field can create a black hole. A black hole made entirely of trapped photons. So light even has gravity.

So energy has mass in that it has inertia and generates a gravitational field.

This is correct. And you don't even need to go to hypothetical examples. The mass of protons and neutrons mostly come from the kinetic energy and potential energy of the constituent quarks and gluons. The mass of the quarks only account for 1% of the proton (and neutron) mass.

I don't think your reply justifies your conclusion - if you'd written "photons have mass in that they have inertia and generate a gravitational field" you'd be set, but this is not what OP is talking about. OP is talking about electrons with higher energy having slightly higher mass as a result. Like ThesiusRisen, I am not sure this is how E=mc2 works.

Electrons gain and lose energy through absorbing and emitting photons right? If a high energy electron and low energy electron had the same "mass/energy", that would seem to imply that photons have no "mass/energy".

Well, all that matters is that energy interacts with gravity - in other words, it has weight. And general relativity tells us that's true regardless of whether it's mass or not.

That being said, in modern times we define "mass" as being proportional to the energy of an object in its own rest frame in flat spacetime. You can have cases where part of that mass is actually some kind of energy (kinetic, electromagnetic, etc.) if you look closely enough - for example, if you have light bouncing around inside a closed box, the box behaves in every way as if its mass includes the electromagnetic energy of the light.

Most comments here are assuming that "empty" space on the phone is occupied by 0 bits, and that new data will switch some to 1. In reality, there's no guarantee that those bits are 0, since most operating systems have clever bookkeeping to allow regions of memory to be marked as empty without having to overwrite that entire section with 0 bits.

Your meta question appears to be: does data have mass? And the answer to that question, and in my mind one of the cool things about data, is that it does not.

A knit quilt with intricate, colorful patterns weighs the same as a gray, featureless one. A book filled with random scribblings weighs the same as an equivalent dictionary. We're a species that evolved to match patterns; data is just patterns: human-interpretable arrangements of things.

This should be higher up. Even if a 1 has more mass than a 0 or vice versa, memory being more or less full doesn't mean that it will weigh more or less. A phone with 0GB used could have more 1s than a phone with 60GB used.

Just no, why on earth would you store the most basic "bit" of binary in something four times as large.. computers are made to be efficient, don't speculate on something you can easily read up but neglect too..

Most of these comments are wildly off topic talking about we codes. Hopefully it doesn't need to be explained why qr codes are not used inside a mobile phones hard drive.

I was only stating that on a hardware level a "bit" (a single 1 or 0, a Boolean value, an 8th of a byte) would not be stored in any other way than a the literal "bit" (1 or 0).

Because physics are a thing and every data storage solution at this tiny scale is noisy to some extent. While maybe not to the degree he is describing, it is very likely that most if not all modern flash storage uses ECC to shield it from bit flips or decaying cells. ECC memory exists for the same reason. No amount of "efficiency" will be preferred over basic data integrity. SSDs for example are massively over provisioned (a 400GB SSD you buy might have over 500GB of pure flash storage internally) and use all kinds of cool "inefficient" tricks to mask the constant cell/page failure that is occurring within the drive.

Dropping a Wikipedia link to site my sources here, but there are super efficient ways to verify data integrity in systems. I couldn't speculate how they do it on phones or drives, but you guys should probably start here and see what you dig up.

While not used in typical file system, this could make the storage a bit safer, allowing easy single-bit (per "big" bit) error correction, i.e. 1010 =1, and 1110 would also read as 1. (though, a triplication 111 or 000 would be more efficient).When you save the data as 10110011, you would't be able to (reliably) recover it from 11110011 if error happens.

For example, Manchester encoding which is used in e.g. IR remotes encodes bits as level changes, so one bit is transferred as either 01 or 10.

Another example are CDs, which have a massive amount or error correction going on - special encodings to avoid long sequences of zeroes, error correction codes and interleaving to be able to correct scratches etc.

Or QR codes, which, depending on the encoding, can be read even if half of the data is missing.

Or HDMI, which uses a 8-to-10 encoding that minimizes DC imbalance on the differential pairs.

And so on, there'd literally hundreds if not thousands of encoding schemes that increase the amount of data in some way.

Remember that flash chips have an integrated controller, unless you're reverse-engineering the gate logic you have no way of telling how the data is stored internally - I wouldn't be surprised at all if the internal controller used some kind of erasure code and simply presented the data as unencoded to the outside interface.

For another example, look up Shingled magnetic recording - to quote Wikipedia, "Device-managed SMR devices hide this complexity by managing it in the firmware, presenting an interface like any other hard disk" - in other words, the data that the system sees is not exactly what's stored on the hard drive.

Generally, the point here is one of the basics of coding theory - symbols != physical states. What the computer sees are symbols, but it's up to the hardware to choose how to realize them - and that realization is what matters in this question.

Unless there's some serious requirement for error correction, that is not the case. Flash memory in consumer devices simply have one voltage level per "code." Current devices usually store more than one bit per memory cell, resulting in exponentially more voltage levels, but there's still no extra encoding overhead.

So eg. with 3 bits per cell (TLC NAND), you have 8 voltage levels, and they represent 000, 001, and so on up to 111. Upcoming QLC (4 bits per cell) double that to 16 voltage levels.

Everybody in this thread is saying this and it's not really true for flash storage. Most operating systems today make use of TRIM commands to tell the storage device which sectors are empty and should be erased in the background at the next best opportunity (i.e. when the device is otherwise idle). This is because erasing flash blocks takes significantly more time than writing an already erased block, so getting it out of the way ahead of time is useful to get faster write speeds later.

This is only true of uncompressed data. Good data compression puts the data in an arrangement statistically indistinguishable from randomness. Any place where entropy is still low is a candidate for further compression.

You're right. But technically, the 0s and 1s of digital information are just two different energy states of the storage medium. One is minimal charge, and the other is slightly more. The computer uses one of them as the 0 and the other as the 1. It doesn't matter which is which, as long as it's consistent. The one with the higher charge has an imperceptably higher mass (according to Einsteins mass/energy equivalence formula). If the phone was filled with all of the higher energy state bytes, then the phone would indeed have more mass than if it was filled entirely with the lower energy ones.

Therefore, we can see that information indeed does not intrinsically have "mass". Instead, information is intrinsically "a difference" in mass (or energy), or more accurately "an organized difference". If you can differentiate between two things, then there is information.

You're wrong about the quilt analogy but not because of the dye (seriously guys? The dye???). A solid coloured knit quilt would be a continuous strand, whereas an intricate pattern involving multiple colours has to start and stop creating denser and heavier points. You can also have several strands of yarn at the same time.

This is true in the case you described, but you could make a single color quilt with the exact same patterns and stitchings as the colored quilt and the analogy still holds. The same amount of material, but with no color contrast to provide information.

Yeah you’re actually wrong about all your assertions. TRIM clears errant data from flash after a delete, and stored data in flash has a calculable difference because of the energy levels for arranged electrons.

A quilt knit with gray wool and one knit with coloured wool would have separate masses, as the dyes would be different. It would just be a small difference.

But even if that were true, how can a digital 1 have more mass than a digital 0. They are both the same thing aren't they? A digital digit.

I find this whole thing intuitively untrue. The physical weight of a device with empty storage should change at all. How could it? 1gb of empty space isn't empty, it just isn't organized meaningfully yet.

With NAND, the opposite would be true. Flash erases to 1 and writes 0. When you erase a NAND cell, the hardware fills the cell with that highest possible amount of voltage. Writing a page then reduces that voltage.

Well, consumes to me means continuous power, so no. An empty die would have more electrons in the cells than a fully written one, by about half. But you wouldn't continuously power the NAND. Otherwise it wouldn't be possible to turn off the device off and keep data. I'd also like to clarify my statement above, I was trying to state that writing more data does not make NAND heavier, but lighter. No idea by how much though.

Edit: it does take a lot more energy to erase a block than to write or read.

There is no clear meaning for a "full" or "empty" phone from an Information Science point of view. The least weighing combination of 1s and 0s could stand for a fully filled storage, simply because that might be the exact information that you want to save. Similarly, the highest weighing combination can be interpreted as empty, as the software dealing with storage information doesn't really bother what state the physical storage is in, it just calls it unused and therefore empty. So this is somewhat about perspectives and what you call full and empty.

From a comp science perspective, IF there is a difference in weight between a bit having a value of 1 and a value of 0, the answer is it depends.

The thing about digital storage is the data is not erased when you delete a file, it's only removed form the "index" so to speak. If you have a 60GB file which is just straight 1s for all of its bits, and you delete it, they don't become 0, they stay at 1 but your file system will set the memory addresses as available for new files to be written to. In the same vein you could have a 60GB file filled with straight 0s. That's why when you delete a file there are file recovery tools that can scan your drive and recover them, as long as you haven't overwritten the data with a new file, it's 1s and 0s all are there and can be read.

It doesn't end there, some chips use inverse logic, where a high (voltage) state means a logic 0 and low state is a logic 1. This is the case for NAND flash memories which most if not all modern smartphones use for their internal storage.

The thing about digital storage is the data is not erased when you delete a file, it's only removed form the "index" so to speak. If you have a 60GB file which is just straight 1s for all of its bits, and you delete it, they don't become 0, they stay at 1 but your file system will set the memory addresses as available for new files to be written to. In the same vein you could have a 60GB file filled with straight 0s. That's why when you delete a file there are file recovery tools that can scan your drive and recover them, as long as you haven't overwritten the data with a new file, it's 1s and 0s all are there and can be read.

This applies to mechanical hard drives, but not really to solid-state storage. The TRIM command will label a chunk of data as out of use, and then the storage controller will get rid of it as part of garbage collection. It may take a little while for the controller to get around to it, but the data will not hang around indefinitely as it would on a hard drive.

Blocks on an SSD can't be written to multiple times, without first erasing the entire block in a time consuming process. So SSDs copy each block one writes to, and enqueued the old copy for erasure so that it can be reused.

It's mainly to spread write activity evenly across the whole drive, because otherwise heavily used cells will wear out prematurely (this is not a concern for hard drives, which have practically unlimited write endurance).

The fact that flash memory can only be erased one full block at a time is also a factor; so smaller chunks (pages) of data that needs to be kept has to be written somewhere else before deletion.

These factors combine to cause write amplification, which the controller is designed to minimize. Some controllers also compress the data before it's written to the flash memory (which means both counteracting write amplification and improving the transfer speed).

The storage controller is simply in a better position to manage these factors than the OS, so it uses a Flash Translation Layer to keep track of where the OS thinks data is, and where it actually happens to be at any given time.

Say I have a full 1TB hard drive. I wipe it and install 500GB of data exactly and tell the computer to fill the remaining space “with 0s”. Would this have any effect on it or does it just free up the space for new things to be written over top of it, with the data still technically there?

Furthermore, encrypted drives (eg on any version of iOS released in the last few years), it’s not possible to tell the difference between an empty chunk of storage and one filled with all 1 bits because both would be statistically indistinguishable from randomness. An encrypted drive will be approximately 50% 0 and 50% 1, even if it’s freshly formatted and contains no real data yet.

And all (or at least all high density) NAND flash is 'whitened' with a scrambling or encryption algorithm to limit interference between adjacent bits. So on average, half of the bits will be set regardless of the data written.

I see what you mean: This depends on how the bits are moved from "0" to "1". If they are energized to a higher state, then we have the same "ever so slightly more weight" thing still going on. To my knowledge pretty much all the memory we use does that so I think it would still apply. But if can find memory where the "higher energized state" is "0" instead of "1" then technically you could say that a 0GB phone weighs more than a 60GB phone, but we'd still have the same issue... I'm not sure if you could get around this issue. I think the way computer memory fundamentally works will get you every time.

Edit: In case I wasn't clear, the contents of memory fundamentally have to have an effect on mass if you are using the type of computer memory we currently use. I can't think of a type of memory that doesn't operate this way, nor can I think of one that could operate any other way. The first point of his comment does not still stand.

If you include the act of flipping the bits as part of the problem, then yes, you'll be burning energy.

Well, that would be the case if we were just moving the electron to a space where it is at the exact same energy state, but to my knowledge all flash memory changes alter the energy state of the electron. If that is the case then the weight would change, but in the case of like SRAM you do technically avoid a relative change in weight because you are storing both possible outcomes anyways.

A phone with 0 GB of memory does not necessarily have it's bits all set to zero though. A phone with 0 GB of storage could have every single bit except a few at the beginning set to 1, while a phone could be full of information that is entirely zeros (again except for the bits at the beginning which stores information about the file structure).

No. A full charged phone will weigh more, though imperceptibly so. The increase in energy stored in the battery directly leads to an increase in mass. Think of a cup of coffee, a hot cup does weigh more than a cold one, simply because it contains more energy. This stems from relativity as can be read in this article excerpted below:

If you have absolutely identical objects that have the same weight exactly when they are at the same temperature, then when one object is heated, it will weigh more. This is because the gravitational force depends on the stress energy tensor in general relativity. The stress energy tensor 00 component is the total energy of the body, which includes the rest mass plus the kinetic energy of the object. Temperature differences means that there is a different amount of kinetic energy in the motion of the atoms of the two bodies.

For example, if you start with two identical kilograms of water at 0 Celsius, and if you then heat one of them to 100 Celsius, then the kilogram at 100 Celsius would be heavier by an amount equivalent to 4.6 nanograms of additional water weight

A slightly discharged battery has less chemical energy and thus less mass, although technically that energy will now be in the form of thermal energy or in the different state of the memory so unless the thermal energy dissipates the phone will have the same mass.

Essentially: However much the energy in the phone changes, unless it somehow leaves the phone like through light or heat conduction, the mass of the phone stays the same.

If I write a block of memory from all 1s to all 0s, all of the electrons are simply returning to the battery

No they are not. Phones generally use Solid-State drives for memory (like Flash). In SSDs a 0 or 1 is stored by using quantum tunnelling or hot carrier emission to "strand" charge/electrons on the "floating island" of a floating gate transistor. So leaving electrons behind is exactly how they work.

A follow up point is that a fully charged phone has the same mass as a dead phone, since charging only increases the chemical potential energy of the battery.

Also not true. There's this little equation called E=mc2 . States of different potential energies have different masses. In fact that's WHAT mass is, energy of a bound state.

To be purely pedantic, no. Because nano means 10-9, there is no perceptible change in weight. A change in the energy levels of electrons is virtually undetectable from a weight perspective. We can calculate out the possible differences, but they are in the 10-18 range of a difference.

In order for the question to be relevant, you would need to double the precision of your orders of magnitude. To put it in perspective, you would need to take a particle on the nano level and grow it to human proportions. Then you’d need to measure the difference in nano based weight on the enlarged particle to notice anything.

No. Flash memory is based on NAND or NOR transistor matrices, which, by there very nature, are non-volatile. Main memory like DDR memory is based on capacitors and, due to the "leaky" nature of capacitors, needs to be refreshed hundreds of times a second to prevent data loss. Both are technically forms of RAM because neither has a sequential nature, unlike progressive sectors on a spinning disk.

Mechanical engineer here. Answer is no. Data saved on your phone is stored on the existing infrastructure, so if you keep your phone in a vacuum and download a bunch of movies the mass of your phone remains constant. Similarly, your battery is the same mass whether full or empty, the electrons simply move around from what I understand. You can write and unwrite data to a storage device all day long and the mass should stay the same.

In short, the answer would be yes. But the answer is much longer than that, since it also depends on: "how does your phone store data?" and most importantly "what is data?".

First let's use an example like a hard drive, that uses a magnetic head to record the data bit by bit in the metalic disk, each bit corresponds to a tiny sector of the disk that is magnetized either positive or negative. In this case, the difference of mass from a empty drive and a full drive would be basically none, since there wasn't really a "change of state" in the parts inside of the hard drive (except for the circuits that drive the magnetic head) but the data storage part of it mostly stay with the same number of electrons and overall mass.

Now, if we're talking about solid state data, or data that is recorded in a method that utilizes transistor states, then I'd say that the difference in mass would be less negligible but still it is just extra electrons in the form of stored data, since this time you'd have to "turn on" some transistors in order to store the data. It would also depend on the configuration of the transistors, some my weigh more empty than full if they're p-type transistors.

The difference of weigh would also be so little... I even found this example that was set by a professor, while talking about solid state data:

"Prof John Kubiatowicz, a computer scientist at the University of California, Berkeley, explained that storing new data involves holding electrons in a fixed place. Using Einstein's E=mc² formula, which states that energy and mass are directly related, Prof Kubiatowicz calculated that filling a 4GB Kindle to its storage limit would increase its weight by a billionth of a billionth of a gram, or 0.000000000000000001g."

It would, but the difference would be negligible. For example, you could also calculate your phones thrust from the light leaving the screen, or the weight of the battery charged vs not charged, ect. On that level there’s a lot you could calculate, but in practical terms it is nothing.